What are the solutions to the equation 2p^2-4p-16=0 ?

The solutions to the equation 2p^2-4p-16=0 are p=4,p=-2

Find the zeros of 2p^2-4p-16=0

The zeros of 2p^2-4p-16=0 are p=4,p=-2

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2 p^{2} - 4 p - 16 = 0

p = 4, p = - 2

Solution steps

2 p^{2} - 4 p - 16 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 16 )}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 (- 16) = 12

p_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 \cdot 2}

Separate the solutions

p_{1} = \frac{- ( - 4 ) + 12}{2 \cdot 2}, p_{2} = \frac{- ( - 4 ) - 12}{2 \cdot 2}

p = \frac{- ( - 4 ) + 12}{2 \cdot 2} : 4

p = \frac{- ( - 4 ) - 12}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

p = 4, p = - 2

- 3 (- 5 x + 2) + x - 3 = 39

6 x + \frac{1}{12} - x - \frac{13}{9} = 5 x - \frac{3}{2} + \frac{x}{30}

\frac{3 - i}{2 + i}

8 x^{2} - 8 x + 2

x + 8 < 0

What are the solutions to the equation 2p^2-4p-16=0 ?
The solutions to the equation 2p^2-4p-16=0 are p=4,p=-2
Find the zeros of 2p^2-4p-16=0
The zeros of 2p^2-4p-16=0 are p=4,p=-2